If $\alpha \neq \beta$ but $\alpha ^ { 2 } = 5 \alpha - 3$ and $\beta ^ { 2 } = 5 \beta - 3$ then the equation whose roots are $\alpha / \beta$ and $\beta / \alpha$ is
[1] $3 x ^ { 2 } - 25 x + 3 = 0$
[2] $x ^ { 2 } + 5 x - 3 = 0$
[3] $x ^ { 2 } - 5 x + 3 = 0$
[4] $3 x ^ { 2 } - 19 x + 3 = 0$
We need the equation whose roots are $\frac { \alpha } { \beta }$ and $\frac { \beta } { \alpha }$ which are reciprocal of each other, which means product of roots is $\frac { \alpha } { \beta } \frac { \beta } { \alpha } = 1 .$ In our choice (a) and (d) have product of roots 1, so choices (b) and (d) are out of court. In the problem choice, None of these is not given. If out of four choices only one of choice satisfies that product of root is 1 then you select that choice for correct answer.
Now for proper choice we proceed as,$\alpha \neq \beta ,$ but $\alpha ^ { 2 } = 5 \alpha - 3$ and $\beta ^ { 2 } = 5 \beta - 3$
changing $\alpha , \beta$ by $x$
Therefore, $\alpha , \beta$ are roots of $x ^ { 2 } - 5 x + 3 = 0$
$\Rightarrow \alpha + \beta = 5 , \alpha \beta = 3$
now, $S = \frac { \alpha } { \beta } + \frac { \beta } { \alpha } = \frac { \alpha ^ { 2 } + \beta ^ { 2 } } { \alpha \beta } = \frac { 19 } { 3 }$ and product
$\frac { \alpha } { \beta } \cdot \frac { \beta } { \alpha } = 1$
Therefore Required equation,
$x ^ { 2 } - ($ sum of roots $) x +$ product of roots $= 0$
$\Rightarrow x ^ { 2 } - \frac { 19 } { 3 } x + 1 = 0$
$\Rightarrow 3 x ^ { 2 } - 19 x + 3 = 0$ is correct answer.
Difference between the corresponding roots of $x ^ { 2 } + a x + b = 0$ and $x ^ { 2 } + b x + a = 0$ is same and $a \neq b ,$ then
[1] $a + b + 4 = 0$
[2] $a + b - 4 = 0$
[3] $a - b - 4 = 0$
[4] $a - b + 4 = 0$
Let $\alpha , \beta$ are roots of $x ^ { 2 } + b x + a = 0$
Therefore $\alpha + \beta = - b$ and $\alpha \beta = a$
again let $\gamma , \delta$ are roots of $x ^ { 2 } + a x + b = 0$
Therefore $\quad \gamma + \delta = - a$ and $\gamma \delta = b$
Now given
$\alpha - \beta = \gamma - \delta$
$\Rightarrow ( \alpha - \beta ) ^ { 2 } = ( \gamma - \delta ) ^ { 2 }$
$\Rightarrow ( \alpha + \beta ) ^ { 2 } - 4 \alpha \beta = ( \gamma + \delta ) ^ { 2 } - 4 \gamma \delta$
$\Rightarrow b ^ { 2 } - 4 a = a ^ { 2 } - 4 b$
$\Rightarrow b ^ { 2 } - a ^ { 2 } = - 4 ( b - a )$
$\Rightarrow ( b - a ) ( b + a + 4 ) = 0$
$\Rightarrow b + a + 4 = 0$ as $( a \neq b )$
If p and q are the roots of the equation $x ^ { 2 } + p x + q = 0 ,$ then
[1] $p = 1 , q = - 2$
[2] $p = 0 , q = 1$
[3] $p = - 2 , q = 0$
[4] $p = - 2 , q = 1$
Given $S = p + q = - p$ and product $p q = q$
$\Rightarrow q ( p - 1 ) = 0$
$\Rightarrow q = 0 , p = 1$
Now If $q = 0$ then $p = 0 \Rightarrow p = q$
If $p = 1 ,$ then $p + q = - p$
$q = - 2 p$
$q = - 2 ( 1 )$
$q = - 2$
$\Rightarrow p = 1$ and $q = - 2$
If a , b , c are distinct positive real numbers and $a ^ { 2 } + b ^ { 2 } + c ^ { 2 } = 1$ then $a b + b c + c a$ is
[1] less than 1
[2] equal to 1
[3] greater than 1
[4] any real no
In such type of problem if sum of the squares of number is known and we needed product of numbers taken two at a time or needed range of the product of numbers taken two at a time. We start square of the sum of the numbers like
$( a + b + c ) ^ { 2 } = a ^ { 2 } + b ^ { 2 } + c ^ { 2 } + 2 ( a b + b c + c a )$
$\Rightarrow 2 ( a b + b c + c a ) = ( a + b + c ) ^ { 2 } - \left( a ^ { 2 } + b ^ { 2 } + c ^ { 2 } \right)$
$\Rightarrow a b + b c + c a = \frac { ( a + b + c ) ^ { 2 } - 1 } { 2 } < 1$
The value of a for which one root of the quadratic equation $\left( a ^ { 2 } - 5 a + 3 \right) x ^ { 2 } + ( 3 a - 1 ) x + 2 = 0$ is twice as large as the other is
[1] -2/3
[2] 1/3
[3] -1/3
[4] 2/3
Let $\alpha , 2 \alpha$ are roots of the given equation therefore sum of the roots
$\alpha +2\alpha =3\alpha =\frac{1-3a}{{{a}^{2}}-5a+3}\ldots (1)$
and product of roots
$\alpha ( 2 \alpha ) = 2 \alpha ^ { 2 } = \frac { 2 } { a ^ { 2 } - 5 a + 3 }\cdots(2)$
By (i) and (ii) we have
$\frac { 9 \alpha ^ { 2 } } { 2 \alpha ^ { 2 } } = \frac { ( 1 - 3 a ) ^ { 2 } } { \left( a ^ { 2 } - 5 a + 3 \right) ^ { 2 } } \times \frac { a ^ { 2 } - 5 a + 3 } { 2 }$
$\Rightarrow 9 \left( a ^ { 2 } - 5 a + 3 \right) = ( 1 - 3 a ) ^ { 2 }$
$\Rightarrow a = \frac { 2 } { 3 }$
If the sum of the roots of the quadratic equation $a x ^ { 2 } + b x + c = 0$ is equal to the sum of the squares of their reciprocals, then $\frac { a } { c } , \frac { b } { a }$ and $\frac { c } { b }$ are in
[1] geometric progression
[2] harmonic progression
[3] arithmetic-geometric progression
[4] arithmetic progression
Given $\alpha + \beta$
$= \frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } } = \frac { ( \alpha + \beta ) ^ { 2 } - 2 \alpha \beta } { \alpha ^ { 2 } \beta ^ { 2 } }$
$\Rightarrow 2 a ^ { 2 } c = b c ^ { 2 } + a b ^ { 2 }$
$\Rightarrow \frac { 2 a } { b } = \frac { c } { a } + \frac { b } { c }$
$\Rightarrow \frac { c } { a } , \frac { a } { b } , \frac { b } { c } \in \mathrm { A.P }$
$\Rightarrow$ reciprocals are in H.P
The number of real solutions of the equation $x ^ { 2 } - 3 | x | + 2 = 0$ is
[1] 4
[2] 1
[3] 3
[4] 2
Given $x ^ { 2 } - 3 | x | + 2 = 0$
If $\quad x \geq 0$ i.e. $| x | = x$
Therefore, the given equation can be written as
$x ^ { 2 } - 3 x + 2 = 0$
$\Rightarrow \quad ( x - 1 ) ( x - 2 ) = 0$
$\Rightarrow \quad x = 1,2$
Similarly for $x < 0 , x ^ { 2 } - 3 | x | + 2 = 0$
$\Rightarrow \quad x ^ { 2 } + 3 x + 2 = 0$
$\Rightarrow \quad x = - 1 , - 2$
Hence $1 , - 1,2 , - 2$ are four solutions of the given equation.
Let two numbers have arithmetic mean 9 and geometric mean 4 . Then these numbers are the roots of the quadratic equation
[1] $x ^ { 2 } + 18 x - 16 = 0$
[2] $x ^ { 2 } - 18 x + 16 = 0$
[3] $x ^ { 2 } + 18 x + 16 = 0$
[4] $x ^ { 2 } - 18 x - 16 = 0$
Let the two number be $\alpha , \beta$
Therefore, $\frac { \alpha + \beta } { 2 } = 9$ and $\sqrt { \alpha \beta } = 4$
Therefore, Required equation
$x ^ { 2 } - 2 ($ Average value of $\alpha , \beta ) x + \sqrt { G M } ^ { 2 } = 0$
$x ^ { 2 } - 2 ( 9 ) x + 16 = 0$
If $( 1 - p )$ is a root of quadratic equation $x ^ { 2 } + p x + ( 1 - p ) = 0$ then its roots are
[1] 0, -1
[2] -1, 1
[3] 0, 1
[4] -1, 2
$1 - p$ is root of $x ^ { 2 } + p x + 1 - p = 0$
$\Rightarrow ( 1 - p ) ^ { 2 } + p ( 1 - p ) + ( 1 - p ) = 0$
$( 1 - p ) [ 1 - p + p + 1 ] = 0$
$\Rightarrow \quad p = 1$
Given equation becomes $x ^ { 2 } + x = 0$
$\Rightarrow \quad x = 0 , - 1$
If one root of the equation $x ^ { 2 } + p x + 12 = 0$ is 4 while the equation $x ^ { 2 } + p x + q = 0$ has equal roots, then the value of q is
[1] 3
[2] 12
[3] 49/4
[4] 4
As $x ^ { 2 } + p x + q = 0$ has equal roots $: p ^ { 2 } = 4 q$
and one root of $x ^ { 2 } + p x + 12 = 0$ is 4
$16 + 4 p + 12 = 0$
OR $p = - 7$
$p ^ { 2 } = 4 q \Rightarrow q = \frac { 49 } { 4 }$
If the roots of the equation $x ^ { 2 } - b x + c = 0$ be two consecutive integers, then $b ^ { 2 } - 4 c$ equals
[1] 3
[2] -2
[3] 1
[4] 2
Let $\alpha , \alpha + 1$ are consecutive integer
Therefore, $( x + \alpha ) ( x + \alpha + 1 ) = x ^ { 2 } - b x + c$ comparing both sides we get $\Rightarrow - b = 2 \alpha + 1$
$c = \alpha ^ { 2 } + \alpha$
$b ^ { 2 } - 4 c = ( 2 \alpha + 1 ) ^ { 2 } - 4 \left( \alpha ^ { 2 } + \alpha \right) = 1$
If both the roots of the quadratic equation $x ^ { 2 } - 2 k x + k ^ { 2 } + k - 5 = 0$ are less than 5 then k lies in the interval
[1] $( 6 , \infty )$
[2] $( 5,6 ]$
[3] [ 4,5 ]
[4] $( - \infty , 4 )$
Given $x ^ { 2 } - 2 k x + k ^ { 2 } + k - 5 = 0$
Roots are less than $5 \Rightarrow D \geq 0$
$\Rightarrow ( - 2 k ) ^ { 2 } \geq 4 \left( k ^ { 2 } + k - 5 \right) \Rightarrow k \leq 5\cdots(A)$
Again $f ( 5 ) > 0$
$\Rightarrow 25 - 10 k + k ^ { 2 } + k - 5 > 0$
$\Rightarrow k ^ { 2 } - 9 k + 20 > 0 \Rightarrow ( k - 4 ) ( k - 5 ) > 0$
$\Rightarrow k < 4 \cup k > 5\cdots(B)$
Also $\frac { \text { sum of roots } } { 2 } < 5 \Rightarrow k < 5\cdots(C)$
from $(\text{A}),(\text{B}),(\text{C})$ we have
$k \in ( - \infty , 4 )$ as the choice gives number $k < 5$
If the equation $a _ { n } x ^ { n } + a _ { n - 1 } x ^ { n - 1 } + \ldots + a _ { 1 } x = 0$ $a _ { 1 } \neq 0 , n \geq 2 ,$ has a positive root $x = \alpha$ , then the equation $n a _ { n } x ^ { n - 1 } + ( n - 1 ) a _ { n - 1 } x ^ { n - 2 } + \ldots + a _ { 1 } = 0$ has a positive root, which is
[1] smaller than $\alpha$
[2] greater than $\alpha$
[3] equal to $\alpha$
[4] greater than or equal to $\alpha$
If possible say
$f ( x ) = a _ { 0 } x ^ { n } + a _ { 1 } x ^ { n - 1 } + \ldots + a _ { n } x$
therefore f ( 0 ) = 0
Now $f ( \alpha ) = 0$ ( therefore x = $\alpha$ is root of given equation )
Therefore, $f ^ { \prime } ( x ) = n a _ { n } x ^ { n - 1 } + ( n - 1 ) a _ { n - 1 } x ^ { n - 2 } + \ldots + a _ { 1 } = 0$
has at least one root in $] 0 , \alpha [$
$\Rightarrow n a _ { n } x ^ { n - 1 } + ( n - 1 ) a _ { n - 1 } x ^ { n - 2 } + \ldots + a _ { 1 } = 0$
has a +root smaller than $\alpha .$
All the values of $m$ for which both roots of the equation $x ^ { 2 } - 2 m x + m ^ { 2 } - 1 = 0$ are greater than - 2 but less than 4 , lie in the interval
[1] $- 2 < m < 0$
[2] $m > 3$
[3] $- 1 < m < 3$
[4] $1 < m < 4$
Let $\alpha , \beta$ are roots of the equation $\left( x ^ { 2 } - 2 m x + m ^ { 2 } \right) = 1$
$\Rightarrow \quad x = m \pm 1 = m + 1 , m - 1$
Now $- 2 < m + 1 < 4\cdots(1)$
and $\quad - 2 < m - 1 < 4\cdots(2)$
$\left\{ \begin{array} { l l } { \Rightarrow } & { - 3 < m < 3 } \cdots(A)\\ { \text { and } } & { - 1 < m < 5 } \cdots(B)\end{array} \right.$
By (A) and (B) we get $- 1 < \mathrm { m } < 3$ as shown by the number line.
If the difference between the roots of the equation $x ^ { 2 } + a x + 1 = 0$ is less than $\sqrt { 5 }$, then the set of possible values of a is
[1] $( 3 , \infty )$
[2] $( - \infty , - 3 )$
[3] $( - 3,3 )$
[4] $( - 3 , \infty )$
$x ^ { 2 } + a x + 1 = 0$
Let roots be $\alpha$ and $\beta ,$ then $\alpha + \beta = - a$ and $\alpha \beta = 1$
$| \alpha - \beta | = \sqrt { ( \alpha + \beta ) ^ { 2 } - 4 \alpha \beta } , | \alpha - \beta | = \sqrt { a ^ { 2 } - 4 }$
since, $| \alpha - \beta | < \sqrt { 5 } \Rightarrow \sqrt { a ^ { 2 } - 4 } < \sqrt { 5 }$
$\Rightarrow \quad a ^ { 2 } - 4 < 5 \Rightarrow a ^ { 2 } < 9 \Rightarrow - 3 < a < 3$
The quadratic equations $x ^ { 2 } - 6 x + a = 0$ and $x ^ { 2 } - c x + 6 = 0$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3 . Then the common root is
[1] 2
[2] 1
[3] 4
[4] 3
Let $\alpha$ and 4$\beta$ be the root of
$x ^ { 2 } - 6 x + a = 0$
and $\alpha$ and 3$\beta$ be those of the equation
$x ^ { 2 } - c x + 6 = 0$
From the relation between roots and coefficients
$\alpha + 4 \beta = 6$ and $4 \alpha \beta = a$
$\alpha + 3 \beta = c$ and $3 \alpha \beta = 6$
we obtain $\alpha \beta = 2$ giving $a = 8$
The first equation is $x ^ { 2 } - 6 x + 8 = 0 \Rightarrow x = 2,4$
For $\alpha = 4,4 \beta = 2 \Rightarrow 3 \beta = 3 / 2$ (not an integer)
So the common root is $\alpha = 2$
